topological vector lattice
A topological vector lattice over is
a vector lattice, and
A topological vector lattice is a topological lattice.
Before proving this, we show the following equivalence on the continuity of various operations on a vector lattice that is also a topological vector space.
. If is continuous, then is continuous too, as and are both continuous under a topological vector space. This proof works in reverse too. , , and are obvious. To see , we see that , since is continuous, is continuous also, so that is continuous. To see , we use the identity , so that , which implies is continuous. Finally, is given by , which is continuous. ∎
Let be a vector lattice. Then for any .
. Next, so that . Since and are both in the positive cone of , so is their sum, so that , which means that . Similarly, . Combining these two inequalities, we see that . ∎
We are now ready to prove the main assertion.
To show that is a topological lattice, we need to show that the lattice operations meet and join are continuous, which, by Lemma 1, is equivalent in showing, say, that is continuous. Suppose is a neighborhood base of 0 consisting of solid sets. We prove that is continuous. This amounts to showing that if is close to , then is close to , which is the same as saying that if is in a solid neighborhood of (), then so is in . Since , . But by Lemma 2, and is solid, as well, and therefore is continuous. ∎
As a corollary, we have
A topological vector lattice is an ordered topological vector space.
|Title||topological vector lattice|
|Date of creation||2013-03-22 17:03:51|
|Last modified on||2013-03-22 17:03:51|
|Last modified by||CWoo (3771)|